Free Math Worksheets

- Write each terminating decimal as a reduced fraction. - When is the square root of a reduced fraction rational? - Check whether both the numerator and denominator are perfect squares.

1. \(A = \sqrt{0.64} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8\). 2. \(D = \sqrt{0.0064} = \sqrt{\frac{4}{625}} = \frac{2}{25} = 0.08\). 3. For \(B\), \(6.4 = \frac{32}{5}\). In lowest terms, the numerator and denominator are not both perfect squares, so \(\sqrt{\frac{32}{5}}\) is irrational. 4. For \(C\), \(0.064 = \frac{8}{125}\). In lowest terms, the numerator and denominator are not both perfect squares, so \(\sqrt{\frac{8}{125}}\) is irrational.

a) \(A = 0.8\) and \(D = 0.08\) b) \(B = \sqrt{\frac{32}{5}}\) and \(C = \sqrt{\frac{8}{125}}\) are irrational because the reduced numerator and denominator are not both perfect squares.

- When is the square root of a natural number rational? - What decimal-expansion property identifies rational numbers? - Simplify radical expressions before classifying them. - Recall the fraction definition of a rational number.

1. \(\sqrt{121} = 11\), so it is rational. 2. The decimal \(0.\overline{45}\) repeats, so it is rational. In fact, \(0.\overline{45} = \frac{45}{99} = \frac{5}{11}\). 3. The number \(\pi\) is irrational. 4. \(\sqrt{2}\sqrt{18} = \sqrt{36} = 6\), so the product is rational. 5. \(\frac{22}{7}\) is a quotient of two integers with a nonzero denominator, so it is rational.

a) Rational b) Rational; it is a repeating decimal. c) Irrational d) Rational; \(\sqrt{2}\sqrt{18} = 6\). e) Rational

- Recall the definition of an irrational number. - What happens when a square root is multiplied by itself? - How many counterexamples are needed to disprove a universal statement?

1. To disprove a universal claim, one counterexample is sufficient. 2. Choose the irrational numbers \(\sqrt{2}\) and \(\sqrt{2}\). 3. Their product is \(\sqrt{2}\sqrt{2} = 2\). 4. Since \(2 = \frac{2}{1}\) is rational, the product of two irrational numbers can be rational. Therefore, the claim is false.

The claim is false. A counterexample is \(\sqrt{2}\sqrt{2} = 2\): both factors are irrational, but the product is rational.

- When is the square root of a fraction rational? - What must be true of the numerator and denominator after the fraction is reduced? - Convert each mixed number to an improper fraction first.

1. \(3\frac{1}{16} = \frac{49}{16}\), so \(\sqrt{\frac{49}{16}} = \frac{7}{4}\), which is rational. 2. \(2\frac{14}{25} = \frac{64}{25}\), so \(\sqrt{\frac{64}{25}} = \frac{8}{5}\), which is rational. 3. \(1\frac{4}{5} = \frac{9}{5}\), so \(\sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}\). Since \(\sqrt{5}\) is irrational, this value is irrational. 4. \(7\frac{1}{9} = \frac{64}{9}\), so \(\sqrt{\frac{64}{9}} = \frac{8}{3}\), which is rational.

a) Rational: \(\frac{7}{4}\) b) Rational: \(\frac{8}{5}\) c) Irrational: \(\sqrt{\frac{9}{5}}\) d) Rational: \(\frac{8}{3}\)

- Recall the definition of a rational number. - Look for counterexamples involving perfect squares. - Think about how repeating decimals can be written as fractions. - Test products of simple irrational square roots.

1. Statement a) is true. Every integer \(z\) can be written as \(\frac{z}{1}\). 2. Statement b) is false. For example, \(\sqrt{9} = 3\), which is rational. 3. Statement c) is true. Every repeating decimal can be converted to a fraction of integers; for example, \(0.\overline{3} = \frac{1}{3}\). 4. Statement d) is false. For example, \(\sqrt{2} \cdot \sqrt{2} = 2\), which is rational even though both factors are irrational.

a) True b) False; for example, \(\sqrt{9} = 3\). c) True d) False; for example, \(\sqrt{2} \cdot \sqrt{2} = 2\).

- Square decimal values to compare them with \(7\). - What property separates rational and irrational numbers? - Think about the decimal expansion of an irrational number.

1. Since \(2.6^2 = 6.76\) and \(2.7^2 = 7.29\), \(2.6 < \sqrt{7} < 2.7\). 2. Every terminating decimal is rational because it can be written as a fraction of integers. 3. If the interval process ended with \(\sqrt{7}\) equal to a rational endpoint, then \(\sqrt{7}\) would be rational. 4. But \(\sqrt{7}\) is irrational. Its decimal expansion is nonterminating and nonrepeating. Rational intervals can approximate it as closely as desired, but no terminating rational endpoint equals it exactly.

a) \(2.6 < \sqrt{7} < 2.7\) b) The claim is false. Every terminating decimal endpoint is rational, but \(\sqrt{7}\) is irrational, so no such endpoint can equal \(\sqrt{7}\) exactly.

- What kinds of decimal expansions represent rational numbers? - How can you guarantee that a fixed block repeats forever? - How can a decimal follow a rule but never become periodic?

1. A rational example is formed by repeating the block \(45\): \(0.454545454545\ldots\). The next ten digits are \(4545454545\). The decimal is rational because it is periodic. 2. An irrational example is \(0.451010010001\ldots\), where successive \(1\)s are separated by one zero, then two zeros, then three zeros, and so on. The next ten digits after the given \(45\) are \(1010010001\). 3. Because the gaps between the \(1\)s keep increasing, no fixed digit block repeats forever. The decimal is nonterminating and nonrepeating, so it is irrational.

a) Example: \(0.454545454545\ldots\). The next ten digits are \(4545454545\), and the repeating pattern makes the number rational. b) Example: \(0.451010010001\ldots\). The next ten digits are \(1010010001\). The increasing gaps prevent periodic repetition, so the number is irrational.

- Add decimal places to create a value between the endpoints. - A fixed repeating block produces a rational number. - To create an irrational decimal, use a nonterminating pattern that never becomes periodic.

1. A terminating example is \(0.455\). 2. A purely repeating example is \(0.\overline{455} = \frac{455}{999}\), which is approximately \(0.455455\ldots\) and lies between \(0.45\) and \(0.46\). 3. For \(\frac{n}{200}\), solve \(0.45 < \frac{n}{200} < 0.46\). Multiplying by \(200\) gives \(90 < n < 92\), so \(n = 91\). Thus, \(\frac{91}{200} = 0.455\). 4. An irrational example is \(0.4501001000100001\ldots\), where the number of zeros between successive \(1\)s increases. This decimal is nonterminating and nonrepeating.

a) \(0.455\) b) \(0.\overline{455}\) c) \(\frac{91}{200}\) d) \(0.4501001000100001\ldots\), with an increasing number of zeros between successive \(1\)s

- Simplify the sum before classifying it. - Simplify the difference before substituting bounds. - Track how multiplying and subtracting change both interval endpoints. - Which decimal digits are fixed throughout the final interval?

1. The sum is \(s = \sqrt{2} + (2-\sqrt{2}) = 2\), which is rational. 2. The difference is \(d = \sqrt{2}-(2-\sqrt{2}) = 2\sqrt{2}-2\). 3. Since \(1.4142 < \sqrt{2} < 1.4143\), multiplying by \(2\) gives \(2.8284 < 2\sqrt{2} < 2.8286\). 4. Subtracting \(2\) gives \(0.8284 < d < 0.8286\). Throughout this interval, the first three digits after the decimal point are \(828\).

a) \(s = 2\), so the sum is rational. b) The first three digits after the decimal point are \(828\).

- First choose a simple rational square root with exactly one decimal place, and square it. - Write a terminating rational number as a reduced fraction and examine the prime factors of its denominator. - How does squaring affect the number of decimal places in a terminating decimal?

1. Choose, for example, \(\sqrt{z} = 0.5\). Then \(z = 0.5^2 = 0.25\), and \(0.01 < 0.25 < 1\). 2. Write a rational number with a terminating decimal as a reduced fraction. Its denominator contains only factors of \(2\) and \(5\). Squaring doubles the exponents of those denominator factors, so the reduced square has an even number of decimal places when unnecessary trailing zeros are removed. 3. The decimal \(0.1\) has one decimal place, which is odd. Therefore, it cannot be the square of a rational number, so \(\sqrt{0.1}\) is irrational.

a) One example is \(z = 0.25\), because \(\sqrt{0.25} = 0.5\). b) It must have an even number of decimal places when written without unnecessary trailing zeros. c) \(\sqrt{0.1}\) is irrational.

- Track how each block length changes. - A decimal can follow a rule without being periodic. - Add the numbers digit by digit and look for a pattern. - Recall how to convert a purely repeating decimal to a fraction.

1. For \(x\), the decimal is built from blocks containing \(n\) consecutive \(1\)s followed by one \(0\), for \(n = 1, 2, 3, \ldots\). 2. For \(y\), the decimal is built from blocks containing \(n\) consecutive \(3\)s followed by one \(4\), for \(n = 1, 2, 3, \ldots\). 3. The block lengths keep increasing, so neither decimal can become periodic. Both decimals are nonterminating and nonrepeating, so both numbers are irrational. 4. At every decimal place, the aligned digits are either \(1+3\) or \(0+4\). Thus, \(x+y = 0.4444\ldots = 0.\overline{4}\). 5. Since \(0.\overline{4} = \frac{4}{9}\), \(x+y = \frac{4}{9}\).

a) In \(x\), a block of \(n\) ones is followed by a zero. In \(y\), a block of \(n\) threes is followed by a four. In both patterns, \(n\) increases by \(1\) each time. b) Both numbers are irrational because their decimal expansions are nonterminating and nonrepeating. c) \(x+y = 0.\overline{4} = \frac{4}{9}\)

- Rational decimals terminate or eventually repeat. - Examine the appended numbers \(10\), \(100\), \(1000\), and so on. - Can a fixed repeating block produce arbitrarily long strings of zeros?

1. A real number is rational exactly when its decimal expansion terminates or eventually repeats. 2. The decimal for \(a\) does not terminate because natural numbers continue to be appended forever. 3. It also cannot eventually repeat. For every natural number \(m\), the appended number \(10^m\) contains \(m\) consecutive zeros after its leading \(1\). Thus, the decimal contains arbitrarily long strings of zeros. 4. An eventually repeating decimal with a fixed period cannot contain arbitrarily long strings of zeros unless the repeating period consists only of zeros, which would make the decimal terminate. Neither case applies here. 5. Therefore, the decimal is nonterminating and nonrepeating, so \(a\) is irrational. A predictable rule alone does not make a number rational.

The claim is false. The decimal neither terminates nor eventually repeats. The blocks \(10^m\) create arbitrarily long strings of zeros, which no fixed nonzero repeating period can produce. Therefore, \(a\) is irrational.